Using factorisation method, solve the following quadratic equations.`(2x)/(x-4)+(2x-5)/(x-3)=25/3; x!=4,3`

Explanation

Given equation is
`(2x)/(x-4)+(2x-5)/(x-3)=(25)/(3)`
`implies(2x(x-3)+(x-4)(2x-5))/((x-4)(x-3))=(25)/(3)`
`implies(2x^(2)-6x+2x^(2)-5x-8x+20)/(x^(2)-3x-4x+12)=(25)/(3)`
`implies(4x^(2)-19x+20)/(x^(2)-7x+12)=(25)/(3)`
`implies3(4x^(2)-19x+20)=25(x^(2)-7x+12)`
`implies12x^(2)-57x-60=25x^(2)-175x+300`
`implies25x^(2)-175x+300-12x^(2)+57x-60=0`
`implies13x^(2)=118x+240=0`
`implies13x^(2)-(78+40)x+240=0`
`implies13x^(2)-78x-40x+240=0`
`implies13x(x-6)-40x+(x-6)=0`
`implies(x-6)(13x-40)=0`
`impliesx-6=0or13x-40=0`
`impliesx=6orx=(40)/(13)`
Hence, 6 and `(40)/(13)` are roots of the equation.